/*
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.
*/

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode dummy_l(-1), dummy_r(-1);
        ListNode *ptr_l = &dummy_l, *ptr_r = &dummy_r;
        for (ListNode *cur = head; cur; cur = cur->next) {
            if (cur->val < x) { // left list
                ptr_l->next = cur;
                ptr_l = ptr_l->next;
            } else { // right list
                ptr_r->next = cur;
                ptr_r = ptr_r->next;
            }
        }
        ptr_l->next = dummy_r.next;
        ptr_r->next = NULL;
        return dummy_l.next;
    }
};

